如何在字符串中找到四元组？

Question

我有这个测试挑战，但我绝对是初学者，所以我坚持了下来，任何合适的解决方案将不胜感激！

我们会说字符串中的“四元组”是连续出现四次的字符。打印给定字符串中四元组的数量。四胞胎可以重叠。

提示：注意空格！

输入格式 带有字符串的单行。

输出格式 数字。

示例输入 abcXXXXXabc

示例输出 2

import java.util.*;

public class Quadruples{

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        String s = sc.nextLine();

        char [] c = s.toCharArray();

        int j=0; int k=1; int m=j+2;

        for(int i=0; i<c.length-1; i++){

            if(c[j] != c[k]){

                j++; k++;

            }else{

                for(int l=0; l<c.length-1; l++){

                    if(c[j] == c[m]){

                        m++;

                    }else{   

                        continue;

                    }

                }

            }

        }

    }

}

Answer 1

您可以使用带有 String substring() 方法的 for 循环：&nbsp; &nbsp; Scanner input = new Scanner(System.in);&nbsp; &nbsp; String str = input.nextLine();&nbsp; &nbsp; int count = 0;&nbsp; &nbsp; for (int i = 0; i < str.length()-3; i++)&nbsp; &nbsp; &nbsp; &nbsp; if (str.substring(i, i+2).equals(str.substring(i+2, i+4))) count++;&nbsp; &nbsp; System.out.print(count);

Answer 2

使用 for 循环的非常简单的暴力破解方法：&nbsp; &nbsp; String input = "abcXXXXXabc";&nbsp; &nbsp; // Counter to store the number of quadruples found&nbsp; &nbsp; int quadrupleCounter = 0;&nbsp; &nbsp; // Loop over each character in the input string (skip the last 3 to prevent going out of bounds)&nbsp; &nbsp; for(int i = 0;i < input.length() - 3; i++) {&nbsp; &nbsp; &nbsp; // Get the current char in the string + the next 3&nbsp; &nbsp; &nbsp; char c1 = input.charAt(i);&nbsp; &nbsp; &nbsp; char c2 = input.charAt(i + 1);&nbsp; &nbsp; &nbsp; char c3 = input.charAt(i + 2);&nbsp; &nbsp; &nbsp; char c4 = input.charAt(i + 3);&nbsp; &nbsp; &nbsp; // Check if all 4 characters are equal&nbsp; &nbsp; &nbsp; if(c1 == c2 && c2 == c3 && c3 == c4) {&nbsp; &nbsp; &nbsp; &nbsp; quadrupleCounter++; // Increase the counter&nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; System.out.println(quadrupleCounter);

Answer 3

或者使用正则表达式（只是为了好玩）：&nbsp; &nbsp; Pattern pattern = Pattern.compile("(.)\\1{3}");&nbsp; &nbsp; int quadrupleCounter = 0;&nbsp; &nbsp; Matcher matcher = pattern.matcher(input);&nbsp; &nbsp; if (matcher.find()) {&nbsp; &nbsp; &nbsp; &nbsp; do {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; quadrupleCounter++&nbsp; &nbsp; &nbsp; &nbsp; } while (matcher.find(matcher.start()+1));&nbsp; &nbsp; }

Answer 4

因此，如果我理解正确的话，“aaaabcccccc”意味着这个字符串中有 4 个四元组，例如，正如您所说的，它们可以重叠。所以这意味着 aaaa 是 1，前四个 cccc 是 2，然后从第二个 c 算起还有一个，所以是 3，然后从第三个 c 算起另一个是 4。如果是这样，那么下面的代码应该没问题因此，我们通过将字符串拆分为字符来创建一个 String 'stringArray'。然后我们使用第一个 for 循环（带有 i）来遍历所有字母，然后在该 for 循环中我们创建另一个 for 循环来检查当前字母是否等于接下来的 3 个字母--->如果是这样，然后我们增加 quadCount（我们找到了多少个四元组）。然后当 forloop 完成时，我们打印出 quadCount。您会看到我们检查 j=i+3 是否小于字符串数组的长度（在第二个 forloop 中），因为如果在我们检查的字符之后直到 letterArray 的末尾没有 3 个字符，那么肯定不会再有 3 个类似的。如果注意空格意味着“cc cc”中没有任何四元组，那么下面的代码是正确的。如果这意味着您不应该考虑空格，并且“cc cc”有一个四元组，那么只需首先调用 String 上的 .trim() 方法，如下所示：String quad= "aaaabccc ccc".trim();-->this将从字符串中删除（修剪）所有空格，然后代码就可以了。public static void main(String[] args) {&nbsp; &nbsp; String quad= "aaaabcccccc";&nbsp; &nbsp; int quadCount=0;&nbsp; &nbsp; String[] stringArray= quad.split("");&nbsp; &nbsp; for(int i=0; i<stringArray.length;i++){&nbsp; &nbsp; &nbsp; &nbsp; String current= stringArray[i];&nbsp; &nbsp; &nbsp; &nbsp; int j=i+3;&nbsp; &nbsp; &nbsp; &nbsp; if(j<stringArray.length){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if(current.equals(stringArray[i+1]) && current.equals(stringArray[i+2]) && current.equals(stringArray[i+3]) ){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; quadCount++;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; System.out.println("Quadcounter final is: "+quadCount);&nbsp; &nbsp; }

Answer 5

未经测试，但我认为您可以从现在开始对其进行测试和更正。public class CheckQuadruple {&nbsp; &nbsp; private Boolean checkQuadrupleInString(String str, int i) {&nbsp; &nbsp; &nbsp; &nbsp; if (i + 3 >= str.length())&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return false;&nbsp; &nbsp; &nbsp; &nbsp;if (str.charAt(i) != str.charAt(i + 1))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;return false;&nbsp; &nbsp; &nbsp; &nbsp;if (str.charAt(i + 1) != str.charAt(i + 2))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return false;&nbsp; &nbsp; &nbsp; &nbsp; if (str.charAt(i + 2) != str.charAt(i + 3))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return false;&nbsp; &nbsp; &nbsp; &nbsp; return true;&nbsp; &nbsp; }&nbsp; &nbsp; public int findQuadruple(String str) {&nbsp; &nbsp; &nbsp; &nbsp; int sum = 0;&nbsp; &nbsp; &nbsp; &nbsp; for (int i = 0; i < str.length(); i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (checkQuadrupleInString(str, i))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; sum += 1;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return sum;&nbsp; &nbsp; }}